Objective

In this lesson we will learn the following calculations:

• BOD loading (seeded and unseeded)
• Solids inventory
• Food/Microorganism Ratio (F/M)
• Mean Ceall Residence Time (MCRT)
• Return sludge rate
• Sludge Volume Index (SVI)
• Sludge Density Index (SDI)
• Detention time

Lecture

Activated Sludge Process

So far we have covered attached growth and fixed film types of wastewater treatment. The activated sludge process uses a suspended growth of organisms to remove BOD and suspended solids. As shown below, the process requires an aeration tank and a settling tank. The air is put into the tank through air diffusers located along the bottom of the aeration tank. The turbulence in the water from the air also mixes the microbes with the food. Thus the air being added provides both the needed oxygen and the required mixing to put the microbes and the food together. After the microbes have eaten the food, we can remove them by allowing them to settle by gravity in a clarifier or settling tank. The microbes go to the bottom of the tank and the clean effluent goes over the weirs. The watery sludge mixture of microbes and solids removed from the bottom of the settling tank is called activated sludge.

During the process, the microbes reproduce. We now have more microbes than before we started. If we returned all the microbes to the aeration tank, there wouldn't be enough air for them all, so we get rid of some of the microbes. The sludge we remove is called the waste activated sludge (WAS). The rest of the activated sludge is returned to the aeration tank. This sludge is called the return activated sludge (RAS). The primary effluent (food) is mixed with the return sludge (microbes) in the aeration tank. The primary effluent combined with the return sludge is called the mixed liquor.

A typical activated sludge process is shown below. The primary effluent, or aeration influent, is mixed with the return sludge. The mixed liquor then enters the aeration tank. Air is added to the tank to give oxygen to the microbes and then mix them with the settled sewage. The mixed liquor leaves the aeration tank and goes to the secondary clarifier. Here the activated sludge settles to the bottom and the clear effluent goes over the weirs. The activated sludge taken from the bottom is then split into two different flows. The waste sludge is removed from the unit and the rest of the sludge is returned to the aeration tank.

Just like the trickling filter, the activated sludge process is used to remove BOD. The activated sludge process generally removes more BOD than the trickling filter. Typically the trickling filter will remove 80 to 85% of the influent BOD. The activated sludge process will often remove 85 to 95% of the BOD from the aeration influent. Activated sludge processes may or may not follow primary treatment. The need for primary treatment is determined by the process modification selected for use. All activated sludge systems include a settling tank following the aeration tank.

• temperature
• return rates
• amount of oxygen available
• amount of organic matter available
• pH
• waste rates
• aeration time
• wastewater toxicity

To obtain desired level of performance in an activated sludge system, a proper balance must be maintained between the amount of food (organic matter), organisms (activated sludge) and oxygen (dissolved oxygen). Let's take a look at some of the process control calculations needed to ensure your activated sludge system is operating properly.

Process Control Calculations

As with other wastewater treatment unit processes, process control calculations are important tools used by operators to control and optimize operations. Let's start with how to calculate the BOD loading for the activated sludge process.

BOD Determination

The term "seed" refers to microbes that consume the biodegradable organic matter in samples for measurement of Biochemical Oxygen Demand (BOD). When determining BOD, it is necessary to have a population of microbes that can oxidize, or consume, the biodegradable organic matter present in the sample. If there is too little seed present in the sample, complete consumption of the organic matter may not occur, resulting in inaccurate results. In samples such as influent and effluent waters prior to disinfection, this is not a problem since the sample will contain a sufficient amount of bacteria to do the job. However, in certain sample types, such as some industrial wastes, high temperature wastes, and treated effluent, there is not enough bacterial activity to consume the material that is present. In these cases, seed must be added. Seed is simply a solution that contains a sufficient population of bacteria and is usually provided at the plant. Let's look at how to determine BOD of a seeded and unseeded sample. (*Note: the 300 mL used in both equation is a standard because a BOD bottle is 300 mL.)

If you have enough microbes in your sample you will not need to use a seed for correction and will use the formula:

Example:

A BOD test was done on a 5 mL sample. The initial DO of the sample is 8.3 mg/L. The final DO of the sample is 4.5 mg/L. Determine the BOD in mg/L for this sample.

The BOD test relies on the presence of healthy organisms. If the samples tested contain materials which could kill or injure the microbes the condition must be corrected and healthy, active organisms are added. This process is known as seeding. If this occurs, determine BOD of the sampling using the formula:

Example:

A BOD test was done on a 5 mL sample. It was determined that the initial DO for the sample was 9.2 mg/L and the final DO was 3.8 mg/L. A seed correction of 0.59 mg/L is needed for the sample. Determine the BOD, in mg/L for the sample.

BOD Loading

When calculating BOD or suspended solids loading on an aeration process (or any other treatment process), loading on that process is usually calculated as lb/day with the equation we've already been using:

Loading, lb/day = mg/L x MGD x 8.34 lb/gal

Example:

The BOD concentration of the wastewater entering an aerator is 180 mg/L. If the flow to the aerator is 2.25 MGD, what is the BOD loading in lb/day?

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 180 mg/L x 2.25 MGD x 8.34 lb/gal

BOD, lb/day = 3,377.7 lb/day

Example:

The BOD concentration of the wastewater entering an aerator is 220 mg/L. If the flow to the aerator is 850,000 gpd, what is the BOD loading in lb/day?

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 220 mg/L x 0.85 MGD x 8.34 lb/gal

BOD, lb/day = 1,559.58 lb/day

Example:

The flow to an aeration tank is 3850 gpm. If the BOD concentration is 155 mg/L, how many pounds of BOD are applied to the aeration tank daily?

First you need to convert gpm to MGD:

Now determine the loading on the process:

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 155 mg/L x 5.54 MGD x 8.34 lb/gal

BOD, lb/day = 7,161.56 lb/day

 

Solids Inventory

In the activated sludge process it is important to control the amount of solids under aeration. The suspended solids in an aeration tank are called mixed liquor suspended solids (MLSS). To calculate the pounds of solids in the aeration tank, we need to know the MLSS concentration and the aeration tank volume. MLSS can be calculated using the following equation:

MLSS, lb = MLSS, mg/L x Volume, MG x 8.34 lb/gal

Example:

If the MLSS concentration is 1525 mg/L and the aeration tank volume is 625,000 gallons, how many pounds of suspended solids are in the aeration tank?

MLSS, lb = MLSS, mg/L x Volume, MG x 8.34 lb/gal

MLSS, lb = 1525 mg/L x 0.625 MG x 8.34 lb/gal

MLSS, lb = 7,949.06 lb

 

Example:

If the MLSS concentration is 850 mg/L and the aeration tank volume is 550,000 gallons, how many pounds of suspended solids are in the aeration tank?

MLSS, lb = MLSS, mg/L x Volume, MG x 8.34 lb/gal

MLSS, lb = 850 mg/L x 0.55 MG x 8.34 lb/gal

MLSS, lb = 3,898.95 lb

 

Some calculations need the mixed liquor volatile suspended solids (MLVSS) measurement. That can be determined from the MLSS measurement:

MLVSS = MLSS x % (decimal) volatile matter (VM)

Example:

The aeration tank contains 3250 mg/L of MLSS. Lab tests indicate the MLSS is 65% volatile matter. What is the MLVSS concentration in the aeration tank?

MLVSS, mg/L = MLSS x % (decimal) volatile matter (VM)

MLVSS, mg/L = 3250 x 0.65

MLVSS, mg/L = 2,112.5 mg/L

Food-to-Microorganism Ratio (F/M Ratio)

After the mixed liquor suspended solids (MLSS) has been determined a mixed liquor volatile suspended solids (MLVSS) test may be performed in order to determine the concentration of volatile suspended solids in the aeration basin. MLVSS is critical in determining the operational behavior and biological inventory of the system. The filter used for MLSS testing is ignited at 550°C for 30 minutes. The weight lost on ignition of the solids represents the volatile solids in the sample. Make sure you understand the difference between MLSS and MLVSS. MLSS is the suspended solids in the aeration tank. They include both organics and inorganis. MLVSS, on the other hand, is the volatile portion only, which basically means this is the portion that are microbes.

The number of microorganisms which are used to seed the aeration chamber is carefully controlled and is based on the food to microorganism ratio (F/M ratio). The microbes will most efficiently break down the organic matter in water if they are present in the right proportion. The food value in the F/M ratio for computing loading can be either BOD or COD (Chemical Oxygen Demand). The reason for biosolids production is to convert BOD to bacteria. One advantage of using COD over BOD for analysis of organic load is that COD is more accurate and doesn't take as long to process.

To illustrate this concept, let's consider a similar situation in which you are feeding hamburgers to hungry football players. If you give 100 hamburgers to 10 hungry football players, many of the hamburgers will remain uneaten. In contrast, if you give 10 hamburgers to 100 hungry players, the players will fight over the hamburgers. The hamburgers are likely to get torn to pieces during the fight and will remain uneaten. The perfect hamburger to football player raio is 1-1. If you have 100 hungry football players and 100 hamburgers the players will not fight over the food and will quickly consume all of the hamburgers.

The food to microorganism ratio follows the same concept. If the appropriate food to microorganism ratio is followed, then there will be efficient BOD removal in the aerator. One manufacturer suggests that the best food to microorganism ratio is about 0.6. There are many different types of activated sludge plants and the typical F/M ratio for those processes are shown below.

In order to calculate the proper amount of microbes to be added to the aeration basin, you will need to use the following formula:

The top portion of the formula represents the amount of BOD going into the aeration. The bottom is the amount of microbes inside the aeration tank that will consume the incoming BOD. The F/M ratio determines how many pounds per day of BOD is available for each pound of microbe in the tank. On exams, you may be given the BOD in m/L and the incoming flowrate in MGD. In this case, you will have to use the lb/day formula we have been using to calculate the BOD in lb/day.

Example:

A wastewater treatment plant has 185,000 lbs of MLVSS in the aeration tank. If the primary effluent has a BOD of 185 mg/L and a flowrate of 4.5 MGD, what is the F/M ratio of the tank?

First determine the BOD in lb/day:

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 185 mg/L x 4.5 MGD x 8.34 lb/gal

BOD, lb/day = 6,943.05 lb/day

You are given the amount of microbes ready to consume the incoming BOD. (MLVSS = 185,000 lb)

Determine the F/M ratio:

Sometimes you will have to do a lot of pre-processing before you can enter the measurements into the F/M ratio formula.

Example:

The aeration tank influent BOD is 155 mg/L and the influent flowrate is 2.28 MGD. What is the F/M ratio if the MLVSS is 2250 mg/L and the aeration tank volume is 1.5 MG?

First, let's determine the BOD in lb/day:

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 155 mg/L x 2.28 MGD x 8.34 lb/gal

BOD, lb/day = 2,947.36 lb/day

That's one portion of the equation: BOD, lb/day. Now we need to determine the pounds of MLVSS in the tank:

MLVSS, lb = MLVSS, mg/L x Volume, MG x 8.34 lb/gal

MLVSS, lb = 2250 mg/L x 1.5 MG x 8.34 lb/gal

MLVSS, lb = 28,147.5 lb

Now that you have all the components, determine the F/M ratio:

Mean Cell Residence Time (MCRT)

The mean cell residence time (MCRT) is an average measure of how long the microorganisms remain in contact with the substrate (food source). MCRT is also known as solids retention time (SRT). MCRT is not a mass balance, but rather a measure of how many days microbes are kept in the activated sludge process before being wasted. The treatment system is constantly generating new solids. As a rule of thumbe, you will produce approximately 0.5 lbs of new solids per pound of BOD removed. So, if you remove 100 lbs of BOD, you will produce approximately 50 pounds of new solids. If you fail to remove the new solids produced, your treatment system will suffer in performance. A longer MCRT age yields less sludge production than a younger MCRT age. This is because BOD (food) is used for both staying alive and growing. You can determine MCRT with the following formula:

Think of this as "solids in the aeration system, lb" divided by "solids leaving the system, lb/day". Let's look at an example:

Let's watch a video showing you how to work MCRT.

Example:

Given the following data, calculate the MCRT:

• MLSS = 2250 mg/L
• Aeration tank + clarifier volume = 1.5 MG
• Effluent TSS = 11 mg/L
• Flow = 4.5 MGD
• WAS flow = 0.3 MGD
• Wasted TSS = 4850 mg/L

You will need to do some pre-processing before you can enter the values into the equation. Look at what you have and what you need. Your measurements are in mg/L, we need them in lb/day:

First, convert the total suspended solids in the aeration system. We know the solids is 2250 mg/L. This covers both the aeration tank and the clarifier.

TSS, lb = MLSS, mg/L x Volume, MG x 8.34 lb/gal

TSS, lb = 2250 mg/L x 1.5 MG x 8.34 lb/gal

TSS, lb = 28,147.5 lb

Now we need to know the lb/day of solids wasted, so we need to convert those measurements from mg/L to lb/day:

Effluent TSS, lb/day = TSS, mg/L x Flow, MGD x 8.34 lb/gal

Effluent TSS, lb/day = 11 mg/L x 4.5 MGD x 8.34 lb/gal

Effluent TSS, lb/day = 412.83 lb/day

You will need to use the waste flow instead of the total flow to the plant for this conversion:

Wasted TSS, lb/day = Wasted TSS, mg/L x Flow, MGD x 8.34 lb/gal

Wasted TSS, lb/day = 4850 mg/L x 0.3 MGD x 8.34 lb/gal

Wasted TSS, lb/day = 12,134.7 lb/day

Now plug the values into the equation:

Return Activated Sludge (RAS)

You may be required to calculate how many pounds of microbes need to be added to the aeration basin over the span of a day. You do not have pure microbes available. Instead, you have a mixture of sludge and microbes from the clarifier. This mixture is known as return activated sludge (RAS). To know how much sludge to add back to the aeration basins, you have to calculate the BOD with your flow to achieve the proper food to microorganism ratio. BOD refers to the amount of oxygen that the microbes need to break down the organics. The return sludge is only made up of approximately two percent microbes, so it can take quite a bit of the plant's sludge to seed the aeration basins. Let's determine how many pounds of RAS need to be added to the aeration basin to achieve the optimum F/M ratio of 0.6.

Example:

The plant has a flow of 850,000 gpd. The BOD of the wastewater is 185 mg/L. You want to achieve a F/M ratio of 0.6 for optimum performance. Determine the amount of microbes needed per day.

First we need to determine the BOD in lb/day for the plant:

BOD, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD, lb/day = 185 mg/L x 0.85 MGD x 8.34 lb/gal

BOD, lb/day = 1,311.47 lb/day

Now you have two components of the formula (BOD and the F/M ratio). With only one missing component, we can easily solve for it. Plug in the values you know:

To get MLVSS on one side by itself to solve for it you would multiply both sides by MLVSS, which cancels it out on the right and moves it to the left, leaving you with:

0.6 x MLVSS, lb = 1,311.47 lb/day

Next step to getting MLVSS alone is to divide both sides by 0.6, which cancels it out on the left, and moves it to the right. Remember, what you do to one side, you must do to the other when moving values around.

This means you you would need to use 2,185.78 lbs of return activated sludge (RAS) to properly seed the aeration basin. What if you needed to know the solution amount in gallons? Convert the pounds of RAS to gallons:

2,185.78 lb x (1 gal/8.34 lb) = 262.08 gal

The question set it up that this is 262.08 gallons per day. If you needed to know how much to feed per minute:

(262.08 gal/day) x (1 day/1440 min) = 0.182 gpm

Return Sludge Rate-Solids Balance (Bioreactor)

In order to determine if your system is balanced properly you take into consideration the MLSS, RAS and the flow to the basin. The balance can be determined with the formula:

Even though the formula has mg/L for MLSS, this needs to be converted to MGD as well, so everything will be in the same unit, allowing you to calculate the return sludge rate, in MGD, needed to keep the solids in the process in balance.

Example:

A plant that has a flow of 2.5 MGD has a MLSS of 2850 mg/L. The current RAS is 6,580 mg/L. Determine the return sludge rate-solids balance needed for this plant.

Let's do the pre-processing to get the mg/L measurements into MGD:

MLSS, lb/day = MLSS, mg/L x Flow, MGD x 8.34 lb/gal
MLSS, lb/day = 2850 mg/L x 2.5 MGD x 8.34 lb/gal
MLSS, lb/day = 59,422.5 lb/day

Now convert to gpd, then MGD:

(59,422.5 lb/day) x (8.34 gal/lb) x (1 MG/1,000,000 gal) = 0.496 MGD

Next we need to do the pre-processing for the return activated sludge:

RAS, lb/day = RAS, mg/L x Flow, MGD x 8.34 lb/gal
RAS, lb/day = 6580 mg/L x 2.5 MGD x 8.34 lb/gal
RAS, lb/day = 137,193 lb/day

Now convert to gpd, then MGD:

(137, 193 lb/day) x (8.34 gal/lb) x (1 MG/1,000,000 gal) = 1.14 MGD

Now let's plug the values into the equation, exchanging the mg/L for MGD:

This particular would need a return sludge rate of 1.92 MGD to keep the solids in the process in balance.

Sludge Volume Index (SVI)

Sludge volume index (SVI) calculations will tell you whether the mixed liquor suspended solids (MLSS) in the aeration tank are settling at the right rate, or if they are hindering the performance of your facility. To calculate the SVI, you must first take a sample from the aeration tank. Let the sample settle for 30 minutes before beginning analysis. Analyze the sample and find out the concentration of suspended solids. this will be your MLSS concentration, represented in grams per liter (g/L). Divide the wet volume of the settled sludge (represented in mL/L) by the MLSS value from the last step. This calculation will give you your SVI value (represented in mL/g).

The typical SVI for a system that is operating as it should will be between 50 and 150 mL/g. If your SVI is outside this range, you may need to adjust the levels in your system. Looking at the characteristics of different samples will give you some clues as to what you can expect from your own system.

To increase the SVI, you will need to increase the waste sludge rate. This will result in a slower rate of settling, which in turn will trap more of the suspended solids in the mixed liquor, leading to a clearer effluent. If you need to decrease the SVI, do the reverse: reduce the waste rate. This results in a thicker sludge with heavier particles. As the density increases, so does the rate of settling, making the process more efficient.

The sludge volume index can be determined with the following equation:

Example:

A sample is tested and it is determined the SSV is 420 mL/L and the MLSS is 2850 mg/L. What is the SVI? What does this indicate about the plant?

This one is just plugging in the given values:

This indicates the plant is running in normal operation, with good settling and a low effluent turbidity.

To determine the sludge density index of the sample:

This means the sludge density index of the sample above would be:

Summary

The activated sludge process uses a suspended growth of organisms to remove BOD and suspended solids. As shown below, the process requires an aeration tank and a settling tank. The air is put into the tank through air diffusers located along the bottom of the aeration tank. The turbulence in the water from the air also mixes the microbes with the food. Thus the air being added provides both the needed oxygen and the required mixing to put the microbes and the food together. After the microbes have eaten the food, we can remove them by allowing them to settle by gravity in a clarifier or settling tank. The microbes go to the bottom of the tank and the clean effluent goes over the weirs. The watery sludge mixture of microbes and solids removed from the bottom of the settling tank is called activated sludge.

During the process, the microbes reproduce. We now have more microbes than before we started. If we returned all the microbes to the aeration tank, there wouldn't be enough air for them all, so we get rid of some of the microbes. The sludge we remove is called the waste activated sludge (WAS). The rest of the activated sludge is returned to the aeration tank. This sludge is called the return activated sludge (RAS). The primary effluent (food) is mixed with the return sludge (microbes) in the aeration tank. The primary effluent combined with the return sludge is called the mixed liquor. To obtain desired level of performance in an activated sludge system, a proper balance must be maintained between the amount of food (organic matter), organisms (activated sludge) and oxygen (dissolved oxygen).

Assignment

Complete the math worksheet for this lesson and return to instructor via email, fax or mail.